Proposition 2.1.6.

  1. is normal subgroup of
  2. is open and closed relative to
  3. An element belongs to if and only if For some natural number and some self-adjoint elements
  4. Note first that is closed under multiplication. (Why? See the comment underneath Definition 2.1.2.)
    We must also show that if belongs to then so do and for every . (Normality)
    Let be a continuous path in from to .
    Then and are continuous paths in from to and to respectively. Not exactly sure how continuity is immediate, perhaps since each point in the path is a unitary, we can continuously take inverses and also conjugate by unitaries
    This proves 1.

Let be the set of elements in of form Where and are all self-adjoint elements in .
It follows from 1. and Lemma 2.1.3 (1), that . Easily
Since for every self-adjoint , we see that is a Group.

Take in and with . Then and by 2.1.3 Lemma (3) and its proof for some self-adjoint element . Hence This shows that is open relative to .
The complement is a disjoint union of cosets of the form with . Each of these cosets are homeomorphic to and therefore open relative to , consequently is closed in .
In conclusion, is a non-empty subse of , is closed and open in , and is connected. This implies that .
Therefore 2. and 3. hold.